Question: Pentagon ABCDE has a vertical line of symmetry. What is the $y$-coordinate of vertex C so that the area of the pentagon is 40 square units? [asy]
unitsize(2mm);
defaultpen(linewidth(.7pt)+fontsize(8pt));
dotfactor=4;

pair A=(0,0), B=(0,4), C=(2,16), D=(4,4), E=(4,0);
pair[] dots={A,B,C,D,E};

draw(B--C--D--E);
dot(dots);

axes(Arrows(4));

label("A(0,0)",A,SW);
label("E(4,0)",E,SE);
label("D(4,4)",D,NE);
label("C",C,NE);
label("B(0,4)",B,NW);
[/asy]
Write the area of pentagon $ABCDE$ as sum the areas of square $ABDE$ and triangle $BCD$.  Since square $ABDE$ has area $4^2=16$ square units, triangle $BCD$ has area $40-16=24$ square units.  If $h$ is the $y$-coordinate of point $C$, the height of triangle $BCD$ is $h-4$ units and its base is $4$ units.  Solving $\frac{1}{2}(4)(h-4)=24$, we find $h=\boxed{16}$.